Let's find the image of the point [tex]\((0, 0)\)[/tex] after two reflections, first across the line [tex]\(y = 3\)[/tex] and then across the [tex]\(x\)[/tex]-axis.
1. First Reflection across the line [tex]\( y = 3 \)[/tex]:
When reflecting a point [tex]\((x, y)\)[/tex] across a horizontal line [tex]\( y = k \)[/tex], the rule is: [tex]\[
(x, y) \rightarrow (x, 2k - y)
\][/tex]
For our specific case:
- The point is [tex]\((0, 0)\)[/tex].
- The line of reflection is [tex]\( y = 3 \)[/tex].
- Using the reflection rule: [tex]\[
(0, 0) \rightarrow (0, 2(3) - 0) = (0, 6)
\][/tex]
So, after the first reflection, the point [tex]\((0, 0)\)[/tex] transforms to [tex]\((0, 6)\)[/tex].
2. Second Reflection across the [tex]\( x \)[/tex]-axis:
Reflecting a point [tex]\((x, y)\)[/tex] across the [tex]\( x \)[/tex]-axis involves the rule: [tex]\[
(x, y) \rightarrow (x, -y)
\][/tex]
Applying this rule to our new point [tex]\((0, 6)\)[/tex]:
- The point is [tex]\((0, 6)\)[/tex].
- Reflecting across the [tex]\( x \)[/tex]-axis: [tex]\[
(0, 6) \rightarrow (0, -6)
\][/tex]
So, after the second reflection, the point [tex]\((0, 6)\)[/tex] transforms to [tex]\((0, -6)\)[/tex].
Hence, the image of the point [tex]\((0, 0)\)[/tex] after the two reflections is [tex]\((0, -6)\)[/tex].
Therefore, the final result is [tex]\((0, 6)\)[/tex] after the first reflection and [tex]\((0, -6)\)[/tex] after the second reflection.
